UCF is a major Metropolitan University located in Orlando Florida. UCF is advertising their bachelor in Economics with the statistic that the starting salary of a graduate with a bachelor in economics is $ 48,500 according to Payscale (2013-13). The Director of Institutional Research at UCF is interested in testing this information. She decides to conduct a survey of 50 randomly selected recent graduate economic students. The sample mean is $43,350 and the sample standard deviation is 15,000. Alpha =0.05 Using the given Null Hypothesis and level of significance
A. The null hypotheisis should not be rejected
B. Not enough information is given to answer this question.
C. I have no idea
D. The null hypotheisis should be rejected D Question 34

D. The null hypotheisis should be rejected

Step-by-step explanation:

This is a hypothesis test for the population mean.

The claim is that the starting salary of a graduate with a bachelor in economics significantly differs from $48,500.

Then, the null and alternative hypothesis are:

The significance level is 0.05.

The sample has a size n=50.

The sample mean is M=43350.

As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=15000.

The estimated standard error of the mean is computed using the formula:

Then, we can calculate the t-statistic as:

The degrees of freedom for this sample size are:

This test is a two-tailed test, with 49 degrees of freedom and t=-2.43, so the P-value for this test is calculated as (using a t-table):

As the P-value (0.019) is smaller than the significance level (0.05), the effect is  significant.

The null hypothesis is rejected.

There is enough evidence to support the claim that the starting salary of a graduate with a bachelor in economics significantly differs from $48,500.

15.5

step-by-step explanation:

5

∑

n = 1

substitute n = 1, 2, 3, 4, 5 and sum the terms

= + + + 2² + 2³

= 0.5 + 1 + 2 + 4 + 8

= 15.5